# Question Video: Coordinates of Regular Polygons at the Origin Mathematics

Find the coordinates of the vertices of a regular pentagon centred at the origin with one vertex at (3, 3).

04:27

### Video Transcript

Find the coordinates of the vertices of a regular pentagon centred at the origin with one vertex at three, three.

Since weβre dealing with a pentagon, letβs see how we can link this with the fifth root of a complex number. We know that, on an Argand diagram, the fifth roots of unity form a regular pentagon. That pentagon is inscribed within a unit circle whose centre is the origin. And one of the vertices lies at the point whose cartesian coordinates are one, zero. So letβs consider the cartesian plane to be an Argand diagram containing our regular pentagon.

We can transform this pentagon to a regular pentagon centred at the origin with a vertex at π§ one by multiplying each of the fifth roots of unity by π§ one. And thatβs the equivalent to finding the fifth root of π§ one to the power of five.

Now we could use De Moivreβs theorem or simply recall that the fifth roots of unity are one, π, π squared, π cubed, and π to the power of four, where π is π to the two π by five π. Since one of the vertices of our pentagon lies at the point three, three, which represents the complex number three plus three π, we can say that π§ one is equal to three plus three π.

Now we know that if π§ one is a root to the equation π§ π minus π€ equals zero and one π all the way through to π to the power of π minus one are the πth roots of unity, then the roots of π§ to the power of π minus π€ equals zero are π§ one, π§ one times π, π§ one times π squared, all the way through to π§ one times π to the power of π minus one. So we can find the coordinates of all the vertices of our regular pentagon by multiplying our π§ one by the fifth roots of unity.

Before we do that though, weβll need to write it in exponential form. The modulus of π§ one is the square root of the sum of the squares of the real and imaginary parts. Thatβs the square root of three squared plus three squared, which is three root two. And since both its real and imaginary parts are positive, we know it lies in the first quadrant. So its argument is the arctan of three divided by three, which is π by four. And we can say that π§ one is equal to three root two π to the π by four π.

The rest of the roots and therefore the other vertices of our pentagon will be given by π§ one times π, π§ one times π squared, and all the way through to π§ one times π to the power of four. To find π§ one times π, itβs three root two π to the π by four π times π to the two π by five π. And remember, to multiply complex numbers in exponential form, we multiply their moduli and then we add their arguments. This means our second root is three root two π to the 13π by 20π.

Now since weβre trying to find the coordinates, weβll need to represent this in algebraic form. And to convert from exponential to algebraic form, we first convert it to polar form. Thatβs three root two times cos 13π by 20 plus π sin of 13π by 20. Distributing the parentheses, we see this is the same as three root two cos of 13π by 20 plus three root two π sin of 13π by 20. And we can see that the second vertex of our pentagon will lie at the point with cartesian coordinates three root two cos of 13π by 20, three root two sin of 13π by 20.

We repeat this process with the third vertex. And we subtract two π from the argument so that we can express the argument within the range for the principal argument. And we see that the third solution is three root two π to the negative 19 over 20ππ. Once again, representing this in polar form and distributing the parentheses, we find the coordinates here are three root two cos of negative 19π by 20, three root two sin of negative 19π by 20.

We can repeat this process for π§ one times π cubed and π§ one times π to the power of four. And we find the vertices of the pentagon to lie at the point whose cartesian coordinates are three, three; three root two cos 13π by 20, three root two sin 13 π by 20, three root two cos of negative 19 π by 20, three root two sin of negative 19π by 20. We have three root two cos of negative 11π by 20, three root two sin of negative 11π by 20, and three root two cos of negative three π by 20, three root two sin of negative three π by 20.